Exponential Relations

Section 6.1 Exponential Relations

This section addresses the following topics.

Read and use mathematical models in a technical document

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Communicate results in mathematical notation and in language appropriate to the technical field

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This section covers the following mathematical concepts.

Read and interpret models (critical thinking)

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Use models including linear, quadratic, exponential/logarithmic, and trigonometric (skill)

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Section 3.3 presented different rates at which data changes including linear, quadratic, and exponential. This section presents models that require exponentials and notation so we can perform calculations. The ability to solve equations with exponentials comes in Section 6.4.

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Subsection 6.1.1 Comparing Growth Rates

In colloquial speech exponential is used to mean “very fast”. In this text we are using the term formally which requires a more detailed definition. This section presents examples that illustrate that an exponential relation is very fast in that it grows faster than linear and quadratic relations. To complete the presentation of comparative rates we will also show another relation that is even faster. Thus we can be more precise about how fast something grows.

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Consider the linear, quadratic, and exponential relations in Table 6.1.1, Table 6.1.2, and Table 6.1.3 respectively. Linear data has the same differences (3 in the example below). In contrast the differences for quadratic and exponential data grow as the data grows. Their growth is not the same however. Quadratic differences grow at the same rate (2nd differences are linear, 4 in the example below). In contrast exponential differences grow faster as the data grows faster (indeed at the same rate).

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Another way to look at this, is that for linears and quadratics eventually the data is bigger than the differences. For example 5,8,11 are all bigger than the difference 3 for the linear below. Likewise, the values 8, 18, and 32 are greater than the differences 6, 10, 14 respectively for the quadratic below. For exponential relations, however, the rate of growth of the values is a multiple of the values. The rate of growth increases as quickly as the values do. For example the values 6, 12, 24 are exactly equal to the differences 6, 12, 24. In this example the values are doubling and so are the differences between them. Because of this exponential growth will always outpace linear and quadratic growth eventually.

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Table 6.1.1. Linear Relation

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$n$	0	1	2	3	4	5
$3n+2$	2	5	8	11	14	17
Differences		3	3	3	3	3

Table 6.1.2. Quadratic Relation

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$n$	0	1	2	3	4	5
$2n^2$	0	2	8	18	32	50
Differences		2	6	10	14	18

Table 6.1.3. Exponential Relation

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$n$	0	1	2	3	4	5
$3(2^n)$	3	6	12	24	48	96
Differences		3	6	12	24	48

We can also compare the ratios of consecutive terms of the relations. Because the ratio can be converted to a percent, we can also think of the growth as percent increase. Table 6.1.4 and Table 6.1.5 have the ratios for a linear and a quadratic relation respectively. From the decimal approximation rows we can see that the ratios are decreasing. A closer look shows they are both trending toward one. In contrast in Table 6.1.6 we are reminded that the ratios for an exponential relation are constant. This also implies that exponential relations grow faster than linear or quadrataic relations.

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Table 6.1.4. Linear Relation

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$n$	0	1	2	3	4	5
$3n+2$	2	5	8	11	14	17
Ratios		5/2	8/5	11/8	14/11	17/14
Percent		250%	160%	137.5%	127.27%	121.43%
Percent Increase		150%	60%	37.5%	27.27%	21.43%

Table 6.1.5. Quadratic Relation

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$n$	1	2	3	4	5	6
$2n^2$	2	8	18	32	50	72
Ratios		8/2	18/8	32/18	50/32	72/50
Percent		400%	225%	177.78%	156.25%	144%
Percent Increase		300%	125%	77.78%	56.25%	44%

Table 6.1.6. Exponential Relation

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$n$	0	1	2	3	4	5
$3(2^n)$	3	6	12	24	48	96
Ratios		2	2	2	2	2
Percent		200%	200%	200%	200%	200%
Percent Increase		100%	100%	100%	100%	100%

The comparisons to linear and quadratic might suggest that exponential is the fastest growing relation. It is not. As one example of a faster relation consider Table 6.1.7 which is known as factorial. Interestingly, the ratios grow linearly. This is vaguely like quadratic relations which grow faster than a linear relation because their differences grow linearly. Factorial relations are growing faster than exponential relations because their ratios are growing linearly rather than remaining constant.

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Table 6.1.7. Factorial Relation

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$n$	0	1	2	3	4	5
$n!$	1	1	2	6	24	120
Ratios		1	2	3	4	5

Checkpoint 6.1.8.

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Subsection 6.1.2 Applications

This section presents some simple applications that introduce us to the nature of exponential relations and introduce us to how to write equations for exponential models. The next few examples are used to present the mathematics, they are not a complete presentation of the science.

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Bacteria (any cells) grow by each cell dividing into two, completely functioning cells. The new cells eventually reproduce by dividing in half as well. This means that the population is doubling. Because each cell takes roughly the same amount of time to grow enough to be able to divide, the population will double again when that much time has elapsed. Because we are interested in the idea, we will not round in the next few examples.

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Example 6.1.9.

The bacteria lactobacillus acidophilus is part of turning milk into yogurt. Based on experiments a new generation of bacteria are formed every 70 minutes, that is, the population doubles every 70 minutes.

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Suppose when we start tracking the data there are 4000 cells. After 70 minutes, all of these will have divided into two, so there will be $4000(2)=8000$ cells. After another 70 minutes, all of these 8000 cells will have divided into two, so there will be $8000(2)=16000$ cells. These results are shown in Table 6.1.10.

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If the population grows by the same ratio every 70 minutes, we might ask ourselves if it grows by the same ratio every 35 minutes. Using math from later in this chapter we can construct Table 6.1.11 which shows populations every 35 minutes. We discover that the population does grow by a fixed percent every 35 minutes. If we look a little deeper we realize that we cut the time in half and switched from doubling (times 2) to times $\sqrt{2} \approx 1.4142\text{.}$

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We could produce a table for any amount of time (e.g., every hour) and we would discover that the population grows by the same multiple every time.

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Table 6.1.10. Growth of Lactobacillus Acidophilus

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Minutes	0	70	140	210
Population	4000	8000	16000	32000
Increased by		4000	8000	16000

Table 6.1.11. Growth of Lactobacillus Acidophilus

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Minutes	0	35	70	105	140
Population	4000	5657	8000	11314	16000
Ratio		1.4142	1.4142	1.4142	1.4142

Having emphasized that exponential data grows by a fixed ratio (multiple) every time unit, we can now work toward mathematical notation.

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Example 6.1.12. First Exponential Model.

To obtain a model (equation) we will review our calculations from the previous example and note a pattern in those calculations.

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The initial amount (information we are given) is $P_0 = 4000\text{.}$ After 70 minutes all of these split into two so the population is $P_{70} = 4000(2) = 8000\text{.}$ After 140 minutes they have all split into two a second time so the population is $P_{140} = 4000(2)(2) = 16000\text{.}$ After 210 minutes they have split a third time increasing the population to $P_{210} = 4000(2)(2)(2) = 32000\text{.}$ We repeatedly mutiply the initial amount by 2. This means we will have a power of 2 times the initial amount.

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The number of 2’s by which we multiply is determined by how many multiples of 70 minutes have expired. $140/70 = 2$ so we multiplied by $2^2\text{.}$ $210/70=3$ so we multiplied by $2^3\text{.}$ In general we multiply by two $t/70$ times.

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Putting these together implies we want to multiply the initial amount (4000 in this case) by two (double the population) for each of multiple of 70. This gives us

\begin{equation*} P = 4000 \cdot 2^{t/70}\text{.} \end{equation*}

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Example 6.1.13. First Exponential Model Redux.

Of course we cannot actually count the number of cells in a colony of bacteria. It is easier to measure by mass (units of grams). If the number of cells has doubled then the mass will have about doubled as well.

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Suppose 3 grams of lactobacillus acidophilus is placed in milk. What is a model for the mass of bacteria if it doubles every 70 minutes?

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From the previous example we know we can multiply the initial amount (3 g) by 2 for each 70 minutes. This gives us

\begin{equation*} A = 3(2^{t/70}). \end{equation*}

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Some exponential relations show the amount decreasing. That is the ratio that is multiplied is between 0 and 1. One example is radioactive decay. Radioactive substances are not stable. The radiation they give off is the result of the atoms breaking down into other substances. This means that over time the amount of the radioactive substance decreases. It has been shown that this decrease is exponential. The rate of decay is expressed in half-life, the amount of time for the substance to be reduced by half.

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Example 6.1.14.

Barium-133 has a half-life of 10.551 years. This means that afeter 10.551 years only half of the original amount will remain. This is the result of the radioactive isotope breaking down into other substances.

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Suppose that we obtain 7.000 grams of Barium-133. After 10.551 years we will have $7.000 \cdot \frac{1}{2} = 3.500$ grams. Note the 1/2 is an exact number (the measurement error is in the number of years). After 21.102 years there will be $7.000 \cdot \left(\frac{1}{2}\right)^2 = 1.750\text{.}$ As before the amount left after $t$ years can be written as the initial amount times the ratio for each 10.551 years.

\begin{equation*} A = 7.000\left(\frac{1}{2}\right)^{t/10.551}. \end{equation*}

Using a algebraic notational change ($2^{-1}=1/2$), this could also be written as

\begin{align*} A \amp = 7.000 \cdot \left(\frac{1}{2}\right)^{t/10.551}\\ \amp = 7.000 \cdot \left(2^{-1}\right)^{t/10.551} \\ \amp = 7.000 \cdot 2^{-t/10.551}. \end{align*}

The negative exponent expresses that the process is decreasing, that is, the amount is going down.

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Exponential relations are not all doubling or cutting in half. The ratio can be anything. Going viral on the internet could be (but is not always) an exponential growth of views.

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Example 6.1.15.

A new cat video is posted and 12 people view it the first day. Every 4 days afterward the number of people who see it triples. Write an equation to model the total number of people who have viewed this video.

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To help ourselves figure this out we can calculate the first few days’ results. The 4 days after the video is posted, there will be $12(3)=36$ views. The eighth day, there will be $12(3)(3)=108$ views. This suggests that for each additional four days we multiply the result by 3.

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Thus we need to divide the number of days by 4 to determine how many times it has tripled. For example after 24 days we expect it to triple $24/4=6$ times.

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To calculate the total number of views we multiply the original twelve views by 3 for each time it tripled. Thus the number of people viewing the video is

\begin{equation*} v = 12(3^{d/4}) \end{equation*}

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Checkpoint 6.1.16.

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We learned in Subsection 3.3.2 that salaries increased by a fixed percent each year are exponential in nature. Now we can write a model for this and calculate results.

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Example 6.1.17.

Tien’s initial salary was $52,429.33. He received a 5% raise each year. What should Tien’s salary be entering the sixth year?

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Because the raise is a 5% increase, each year’s salary is 105% of the previous years. In decimal the percent is $p=1.05\text{.}$ Thus after one year his salary will be $\$52,429.33(1.05)\text{.}$ After two years his salary will be $\$52,429.33(1.05)(1.05)\text{.}$ This pattern will continue.

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The model then initial salary time 1.05 for each year.

\begin{equation*} S=\$52,429.33(1.05)^t \end{equation*}

where $t$ is the number of years since he was hired. Entering the sixth year would mean he has just received his fifth raise. His salary would be

\begin{equation*} S=\$52,429.33(1.05)^5 = \$66,914.59. \end{equation*}

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Example 6.1.18.

If Moses’ salary after six raises was $72,311.54, and he received a 4% raise each year. What was his initial salary?

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Because the raise is a 4% increase, the percent is $p=1.04\text{.}$ The model then is each year’s salary is the original salary times 1.04 for each raise.

\begin{equation*} S=S_0(1.04)^t \end{equation*}

where $t$ is the number of years since he was hired, and $S_0$ is the initial salary. We can now solve for the initial salary.

\begin{align*} \$72,311.54 \amp = S_0(1.04)^6.\\ \$72,311.54 \amp \approx S_0(1.2653).\\ \frac{\$72,311.54}{1.2653} \amp \approx \frac{S_0(1.2653)}{1.2653}.\\ \$57,149.72 \amp \approx S_0. \end{align*}

Because the salary would be rounded each year this might be off by a small amount, but not enough to matter for our curiosity.

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In the previous example we rounded in the second step to 4 decimal places. Was this the right choice? We can test by trying the calculation with fewer decimal places. If we had rounded to 5 places, the last step would have been $\frac{\$72,311.54}{1.26532} \approx \$57,148.82\text{.}$ If we had rounded to 6 places, the last step would have been $\frac{\$72,311.54}{1.265319} \approx \$57,148.86\text{.}$ If we had rounded to 7 places, the last step would have been $\frac{\$72,311.54}{1.265320} \approx \$57,148.86\text{.}$ Likewise testing 8 places would not change the amount. Thus any more than 6 places is not enough to change the result. However, was the 4 places result bad? The amounts between the roundings are different from each other but not by enough to impact anyone’s life. Remember rounding is often about being practical: if the variation we round away has no impact, why bother with all the extra work?

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Because we just calculated an initial salary from a current salary and the annual percent increase, we might wonder if we can calculate the percent increase from a past and current salary. We can, but that calculation is not an exponential calculation. Rather it requires a technique for roots from Section 5.3.

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Example 6.1.19.

We will calculate the percent increase given initial and final salaries. If Raven’s initial salary was $53,242.17, and her salary at the end of 7 years was $67,368.33, what was her annual percentage increase?

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Solution.

The end of seven years means there have been six raises. From the two data points we know

\begin{align*} 67368.33 & = 53242.17(r^6)\\ \frac{67368.33}{53242.17} & = r^6\\ \sqrt[6]{\frac{67368.33}{53242.17}} & = \sqrt[6]{r^6}\\ 1.04 & = r \end{align*}

Thus her annual percentage increase was 4%. The full model is

\begin{equation*} S = 53242.17(1.04)^t \end{equation*}

where $t$ is time in years.

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Subsection 6.1.3 Two Exponential Models

Above we worked examples in which some initial amount (e.g., mass of bacteria or salary) was repeatedly multiplied by a number. These were written in slightly different ways giving us the following two forms of the exponential model.

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Model 6.1.20. Exponential Growth (rate).

An amount that is growing exponentially with respect to time is given by

\begin{equation*} A = A_0\left( r^{t/d} \right) \end{equation*}

where

$A$ is the final amount (units vary)

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$A_0$ is the initial amount (same units as $A$)

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$r$ is the doubling ($r=2$), trebling ($r=3$), or other increase. It is unitless.

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$t$ is the amount of time (units typically days/hours/minutes)

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$d$ is how long until the increase occurs (same units as $t$)

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Model 6.1.21. Exponential Growth (percent).

An amount that is growing exponentially with respect to time is given by

\begin{equation*} A = A_0\left(1+p \right)^t \end{equation*}

where

$A$ is the final amount (units vary)

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$A_0$ is the initial amount (same units as $A$)

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$p$ is the percent increase or decrease. It is unitless.

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$t$ is the amount of time (units typically days/hours/minutes)

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It is possible to convert from one form to the other. The following examples demonstrate determining the percent increase/decrease for both forms which may require converting to the second forms. Percent increase and decrease may be reviewed in Subsection 2.1.2.

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Example 6.1.22.

Suppose $P=100(1.032)^t$ where $t$ is the number of years. What is the annual percent increase?

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We know that

\begin{align*} 1.032 \amp = 1+p\\ 1.032-1 \amp = p\\ 0.032 \amp = p \end{align*}

so this is a 3.2% increase.

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Example 6.1.23.

Suppose $P=100(0.88)^t$ where $t$ is the number of years. What is the annual percent decrease?

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We know that

\begin{align*} 0.88 \amp = 1+p\\ 0.88-1 \amp = p\\ -0.12 \amp = p \end{align*}

Because $p$ is negative this is a decrease. The annual percent decrease is 12%.

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Generally if the value $1+p$ is greater than one the exponential is increasing (represents a percent increase), and if the value is less than one, the exponential is decreasing (represents a percent decrease).

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Example 6.1.24.

Suppose $P=230(1.03)^t\text{.}$ Is this increase or decrease and what is the percent increase/decrease?

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Solution.

Because $1.03 > 1.00$ this is a percent increase.

\begin{align*} 1.03 \amp = 1 + p\\ 1.03-1 \amp = p\\ 0.03 \amp = p \end{align*}

Thus this is a 3% increase, because the percent increase/decrease is how much more/less than 100%.

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Even when the exponential model is expressed in terms of doubling or similar, we can determine the percent increase or decrease. This requires knowing the following property $2^{t/3}=(2^{1/3})^t \approx (1.2599)^t$ or in general

\begin{equation*} b^{t/r}=(b^{1/r})^t. \end{equation*}

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Example 6.1.25.

In the exponential model $P=342\left( 2 \right)^{t/2.06}$ what is the percent increase or decrease? Note these numbers are measurements.

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First, we use the property stated above to convert the form

\begin{align*} \left( 2 \right)^{t/2.06} & = \\ \left( 2^{\frac{1}{2.06}} \right)^t & \approx 1.400009768^t\\ & \approx 1.40^t \end{align*}

Because 1.4 is bigger than one it is a percent increase. Note $1.40=1+0.40\text{,}$ so this is a 40% increase, because the percent increase/decrease is how much more/less than 100%.

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For rounding we used that 2 and 1 (in the division) are exact numbers. Calculating the power of 2 maintains the number of significant digits (3). When we subtract precision is maintained to the hundredths position (2 significant digits).

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Significant Digits beyond Arithmetic.

In Significant Digits beyond Arithmetic we noted that a square root maintains the same number of significant digits. The same mathematics applies to exponential. For example, $2^{3.51} \approx 11.39240156 \approx 11.4\text{.}$

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Example 6.1.26.

In the exponential model $P=342\left( 2 \right)^{-t/27}$ what is the percent increase or decrease?

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First, we notice that the exponent is negative which we know means the amount is decreasing. To determine the percent decrease we use the algebra illustrated above to convert the form

\begin{align*} \left( 2 \right)^{-t/27} & = \\ \left( 2^{\frac{-1}{27}} \right)^t & = 0.9746546091^t\\ & \approx (0.97)^t \end{align*}

Because 0.97 is less than one it is a percent decrease. Note $1-0.97=0.03\text{,}$ so this is a 3% decrease, because the percent increase/decrease is how much more/less than 100%.

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We round to two decimal places because 27 has 2 significant digits. The subtraction retains precision to the hundredths position which is only one significant digit.

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Checkpoint 6.1.27.

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A common base for exponentials in many scientific models is $e\text{.}$ We can work with this base using calculation devices which will have an $e^x$ button or an $\text{exp}(x)$ function.

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Example 6.1.28.

What is the percent increase or decrease for the exponential model $P=27e^{0.0200t}\text{?}$

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Just as with the fractions we convert to decimal. Note that we can also write the model $P=27(e^{0.0200})^t\text{.}$ Do you see the difference? $e^{0.0200} \approx 1.02$ which we found using a device. Thus this is a percent increase of 2%.

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